Question: What is the slope of the line tangent to $f(x) = -x^{2}+x+8$ at $x = -3$ ?
The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-(x+\Delta x)^{2}+x+\Delta x+8) - (-x^{2}+x+8)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-(x^{2}+2x \Delta x+\Delta x^{2})+x+\Delta x+8) - (-x^{2}+x+8)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-x^{2}-2(x \Delta x)-\Delta x^{2}+x+\Delta x+8+x^{2}-x-8}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-2(x \Delta x)-\Delta x^{2}+\Delta x}{\Delta x}$ $ = \lim_{\Delta x \to 0} -2x-\Delta x+1$ $ = -2x+1$ $ = (-2)(-3)+1$ $ = 7$